2008COMC加拿大数学公开赛真题答案免费下载

历年 Canadian Open Mathematics Challenge加拿大数学公开赛

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2008 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由两部分题目组成

Part A共8题，每题5分

Part B共4题，每题10分

共计12题，满分80分

不可使用任何计算器

完整版下载链接见文末

部分真题预览：

Part A Solutions:

A7)We make a list of the possible paths that the star could take until it either reaches the top left corner or it moves off the grid. We use “L” to represent a move to the left and “U” to represent a move upwards. The possible paths are:
LLL, LLUL, LLUU, LULL, LULU, LUUL, LUUU,
UUU, UULU, UULL, ULUU, ULUL, ULLU, ULLL
Of these paths, 6 of them reach the top left corner square, namely LLUU, LULU, LUUL, UULL, ULUL, and ULLU.
Each time the star moves, the probability that it moves to the left is 1/2 and the probability that it moves upwards is 1/2 . Thus, the probability that the star follows a particular path of length 3 is (1/2)³= 1/8 and the probability that the star follows a particular path of length 4 is (1/2)⁴= 1/16 . (We can note that we have 2 paths of length 3 and 12 paths of length 4, which is consistent with the probabilities above, because 2(1/8) + 12( 1/16) = 1/4 + 3/4 = 1.)
Since there are 6 paths of length 4 that reach the top left corner, the probability that the star reaches this square is 6( 1/16) = 3/8 .
Answer: 3/8

Part B Solutions:

B3)

1. 1. Setting m = n = 0, the given equation becomes a₀ + a₀ = 1/2a₀ + 1/2a₀ or 2a₀ = a₀ or a₀ = 0.
   2. Setting m = 1 and n = 0, the given equation becomes a₁ + a₁ = 1/2a₂ + 1/2a₀
      Since a₀ = 0 and a₁ = 1, we obtain 1 + 1 = 1/2a₂ + 1/2(0) or 1/2a₂ = 2 or a₂ = 4.
      One way to obtain a₃ in an equation by using values of m and n that are as small as possible is by setting m = 2 and n = 1. In this case, the equation becomes a₁ + a₃ = 1/2a₄ + 1/2a₂ (*)
      Since we already know the values of a₁ and a₂, then (*) requires us to determine a₄ in order to determine a₃.
      We can determine a₄ by setting m = 2 and n = 0, which gives a₂ + a₂ = 1/2a₄ + 1/2a₀ from which we get 4 + 4 = 1/2a₄ + 1/2(0) or 1/2a₄ = 8 or a₄ = 16.
      Substituting back into (*), we get 1 + a₃ = 1/2(16) + 1/2(4) or 1 + a₃ = 8 + 2 or a₃ = 9. Thus, if a₁ = 1, then a₂ = 4 and a₃ = 9. (If we set m = k and n = 0, the given equation would become a_k + a_k = 1/2a_2k + 1/2a₀. Since a₀ = 0, then a_2k = 4a_k, which is a more general statement than a₂ = 4a₁ and a₄ = 4a₂ that were discovered above.)
2. Two Solutions:
   1. Solution 1: Let M be a positive integer.
      Setting m = n = M, the given equation becomes b₀ + b_2M = b_2M + b_2M or b₀ + b_2M = 2b_2M or b₀ = b_2M. This tells us that each even-numbered term in the sequence has the same value.
      Next, set m = M and n = 0. In this case, the given equation becomes b_M + b_M = b_2M + b₀
      Since b_2M = b₀, this equation becomes 2b_M = 2b₀ or b_M = b₀.
      Therefore, b_M = b₀ for every positive integer M.
      Hence, all terms in the sequence have the same value, as required.
   2. Solution 2: Let M be a positive integer.
      Setting m = n = M, the given equation becomes b₀ + b_2M = b_2M + b_2M or b₀ + b_2M = 2b_2M or b₀ = b_2M.
      This tells us that each even-numbered term in the sequence has the same value.
      Next, set m = 1 and n = 0. In this case, the given equation becomes b₁ + b₁ = b₂ + b₀
      Since b₂ = b₀, this equation becomes 2b₁ = 2b₀ or b₁ = b₀.
      Next, setting m = M and n = M − 1, the equation becomes b₁ + b_2M−1 = b_2M + b_2M−2
      Since b₁ = b₀ (by the above) and b_2M = b_2M−2 = b₀ (because these are even-numbered terms), this equation becomes b₀ + b_2M−1 = b₀ + b₀ or b_2M−1 = b₀.
      Therefore, for any positive integer M, we have b_2M = b₀ and b_2M−1 = b₀. In other words, every even-numbered term and every odd-numbered term is equal to b₀.
      Hence, all terms in the sequence have the same value, as required.

 

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